If   ${\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\frac{5\pi }{2}$and  ${\sin }^{-1}\frac{x}{a}-{\sin }^{-1}\frac{y}{b}=\frac{\pi }{12}$then $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is

   $$\begin{gathered} Given {\cos }^{-1}\frac{x}{a}+ {\cos }^{-1}\frac{y}{b}=\frac{5\mathrm{\pi }}{12} ——————\to \left(1\right) \\ and {\sin }^{-1}\frac{x}{a}-{\sin }^{-1}\frac{y}{b}=\frac{\mathrm{\pi }}{12} \\ \Rightarrow \left(\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\frac{x}{a}\right)-\left(\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\frac{y}{b}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow -{\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\frac{\mathrm{\pi }}{12}————————\to \left(2\right) \\ \left(1\right)-\left(2\right)\Rightarrow 2{\cos }^{-1}\frac{x}{a}=\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\cos }^{-1}\frac{x}{a}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow \frac{x}{a}=\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2} \\ Also, \left(1\right)+\left(2\right)\Rightarrow 2{\cos }^{-1}\frac{y}{b}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\cos }^{-1}\frac{y}{b}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{y}{b}=\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}} \\ \therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{5}{4}=1.25 \end{gathered}$$